Question

How do you evaluate the limit `sin(3x)/x` as x approaches `0`?

Answer 1

Use `lim_(theta rarr 0)sin theta /theta = 1`.

Explanation:

One way to use `lim_(theta rarr 0)sin theta /theta = 1` is to use `theta = 3x`

But now we need `3x` in the denominator.

No problem, multiply by `3/3`

`lim_(xrarr0) sin(3x)/x = lim_(xrarr0) 3 * sin(3x)/((3x))`

As `xrarr0`, s also `3x rarr0`. We can substitute to get

`lim_(theta rarr0)3*sin theta/theta = 3*1 = 3`

I like the first method (above) Here's a second method.

`sin(3x) = sin(x+2x) = sinx cos(2x)+cosx sin(2x)`

`= sinx(cos^2x-sin^2x) + cosx(2sinxcosx)`

`= sinx(3cos^2x-sin^2x)`

So

`lim_(xrarr0) sin(3x)/x = lim_(xrarr0)(sinx(3cos^2x-sin^2x))/x`

`= lim_(xrarr0)((sinx/x)(3cos^2x-sin^2x))`

`= (1)(3(1)^2 - 0^2) = 3`